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Find 1/(sin 45 sin 46) + 1/(sin 47 sin 48) + ... + 1/(sin 133 sin 144).

 Oct 25, 2019
 #1
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Ok so i cant help srry

 Oct 25, 2019
 #2
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Find 1/(sin 45 sin 46) + 1/(sin 47 sin 48) + ... + 1/(sin 133 sin 144).


\(=\frac{1}{ (sin 45 sin 46)} + \frac{1}{(sin 47 sin 48) }+ ..         . + \frac{1}{sin (180-133) sin (180-144))}\\ =\frac{1}{ (sin 45 sin 46)} + \frac{1}{(sin 47 sin 48) }+ ..         . + \frac{1}{sin (47) sin (46))}\\ =\frac{1}{ sin 45 sin 46} + \frac{1}{sin 46sin 47} + \frac{1}{sin 47 sin 48 }.....+ \frac{1}{sin89sin90}\\ \)

 

Yea, I know, that does not help much.

 Oct 25, 2019
 #4
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sumfor(n, 45, 144, 1 / (sin n * sin (n+1))) = -1.511943134.....

 

∑[1/ [sin(n)*sin(n+1)], n, 45, 144] =-1.511943134.........

 Oct 25, 2019
edited by Guest  Oct 25, 2019

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